Question

In a uniform magnetic field with induction B directed vertically down , a charged ball with charge Q rotates uniformly horizontally. the Ball is suspended on a thread length L=1 m. The angle of deviation of the thread from the vertical is a. The speed of the ball V, and its mass m.

B=2.7

Q=-2,5*e

a=0

m=1,5me

V=21

Answer 1

Radius of circular path of charge in uniform magnetic field is given by r = mv/Bq where v is the component oof its velocity which is perpendicular to the direction of magnetic field .In this case the plane of motion is horizontal and B is vertical so, vel is always perpendicular to B,Me is the mass of electron = 9.1 * 10^-31

substituting, we get, r=1.5me *21/2.7*(-2.5 *e) = 1.5*9.1*10^-31 *21/(2.7*-2.5 *1.6*10^-19) the -ve sign shows that it moves opposite to right hand thumb rule.

=>r = 26.54 x 10^-12 m now, angle suspended by the thread at wall can be obtained by sin(theta) = opp/hyp

=> sin(a) = r/L = 26.54 *10^-12 => a~= 0